A battery of 10 V is connected in series with resistors of 3 - 5 - 4 - 5 - and 13 - - How much current would flow through the 3 - resistor- -2 Mark-

10 V is connected in series with resistors of 3 Ω, 5 Ω, 4 Ω, 5 Ω and 13 Ω . so, R_eq = 3 + 5 + 4 + 5 + 13 = 30 nbsp;Ω nbsp;[1 Mark] We know from ohms law ...

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Standard X

Physics

Equivalent Resistance of a System of Resistor

A battery of ...

Question

A battery of 10 V is connected in series with resistors of $3 Ω, 5 Ω, 4 Ω, 5 Ω a n d 13 Ω .$ How much current would flow through the $3 Ω$ resistor? [2 Mark]

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Solution

10 V is connected in series with resistors of $3 Ω, 5 Ω, 4 Ω, 5 Ω a n d 13 Ω .$
so,
$R_{e q} = 3 + 5 + 4 + 5 + 13 = 30$ $Ω$ [1 Mark]

We know from ohms law
$I = \frac{V}{R} = \frac{10}{30} = 0.33 A$ [1 Mark]

Same current will flow from all the resistances.

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A battery of 10 V is connected in series with resistors of 3 Ω,5 Ω,4 Ω,5 Ω and 13 Ω. How much current would flow through the 3 Ω resistor? [2 Mark]

10 V is connected in series with resistors of 3 Ω,5 Ω,4 Ω,5 Ω and 13 Ω. so, Req=3+5+4+5+13=30 Ω [1 Mark] We know from ohms law I=VR=1030=0.33A [1 Mark] Same current will flow from all the resistances.

A battery of 10 V is connected in series with resistors of $3 Ω, 5 Ω, 4 Ω, 5 Ω a n d 13 Ω .$ How much current would flow through the $3 Ω$ resistor? [2 Mark]

10 V is connected in series with resistors of $3 Ω, 5 Ω, 4 Ω, 5 Ω a n d 13 Ω .$
so,
$R_{e q} = 3 + 5 + 4 + 5 + 13 = 30$ $Ω$ [1 Mark]

We know from ohms law
$I = \frac{V}{R} = \frac{10}{30} = 0.33 A$ [1 Mark]

Same current will flow from all the resistances.